P^2+Σ2P

I read " BEYOND REASON ". I don't know about Turing Machine, but this is quite interesting because I can still find the similarity with web programing in these days. I also count prime numbers, so I can study from the algorithm of Turing Machine. The realm is completely different. However, it seems to be a new light for me. Turing Machine is like an infinite tape. It read a sign from each cell. For example, there are three cells, such as 0(zero),1,X. 0 1 X A 0NA 1NA XRB B 0RB 1RB 0RC C XLD XLD XLD D 0LD 1LD XNE E Stop Stop Stop This is like a dictionary or code. 0="Put 0 in the cell". 1="Put 1 in the cell". N="Don't move the cell". A="Stay in the same cell". X="Put X in the cell". R="Move the cell to the right". B="The new state". C="The end of the line.Then put X". L="Move the cell to th...

When you see the golden key , π(x) would be related to zeros of Riemann's zeta function. J function is also based on π(x). Therefore, if you counted prime numbers, you would guess zeros of Riemann's zeta function. ζ(s)=1/(1-1/2^s)×1/(1-1/3^s)×1/(1-1/5^s)×・・・ When s is 0.5+zi, ζ(s)=0. Here is a Riemann's zero. s=14.1435657,21.0279853,・・・ In order to use the calculator, you need to put an integer which must be close to 0.5+zi. It is very hard to guess it, because there would be no pattern about it. However, when you calculate J function, you would know that it is almost the same with 0.5+zi. Of course, this isn't perfect. J function is related to prime numbers. The pattern of prime numbers would be fractal, and I can calculate π(x). Therefore, as long as zeros of Riemann's zeta function is connected to π(x), it would move with the same pattern as inflation. I know that the calculation is complicated. x→J(x)→0.5+zi 40→14.61667→14.1435657 65→21.5333→21....

I don't understand what infinity is, but Cantor set is almost similar with prime numbers and my blog. At first, you get rid of the center of each block. This is the infinite pattern, so I can also write this. You keep deleting the center. F0=1 F1=1/3 F2=1/9 F3=1/27 ・ ・ ・ F(x)=1/3^x If x is infinite, F(x) may be Zero, but it isn't. Imaginary lines may be real, if zero exists. Basically, Cantor set is expressed by on and off, but zero would mean none duality of duality. You always have the one block, and you take the center off. Therefore, 0*3^∞ must be 1, and this is ridiculous. Is this the philosophical problem? I'm not so curious about it.

I found the pattern of prime numbers , but I know that it is almost the same with sieve of Eratosthenes . When you see odd numbers, there is symmetry and you can find sieve of Eratosthenes. 3 5 7 9 11 3 9 15 21 27 33 5 15 25 35 45 55 7 21 35 49 63 77 9 27 45 63 81 99 11 33 55 77 99 121 I mean x=(3,5,7,9,11) y=(3,5,7,9,11). Therefore, you can see z=x*y and z=x+Σ2x or z=y+Σ2y. 9=3*3 and 9=3+2*3,15=3*5 and 15=3+4*3 or 15=5*3 and 15=5+2*5, 21=3*7 and 21=3+6*3 or 21=7*3 and 21=7+2*7・・・・. z is not prime numbers, so x'=(3,5,7,11) y'=(3,5,7,11). However, when you add 2, the symmetry is broken. 3 5 7 9 11 2 6 10 14 18 22 3 9 15 21 27 33 5 15 25 35 45 55 7 21 35 49 63 77 9 27 45 63 81 99 11 33 55 77 99 121 x=(3,5,7,9,11)...

Riemann's zeta function is known as contraction. This is also contraction. Then you can get the golden key. Zeros of Riemann's zeta function is related to π(x). Therefore, if you can know the golden key, you may understand zeros of Riemann's zeta function. As long as I use PC to count prime numbers, it seems to be the same with J(x). This is assumption. I know what I don't know.

It is very difficult to see what Riemann's zeta function is. I don't write about 0 and I can't understand it. However, it is connected with prime numbers. Now I got this formula. logζ(s)=1/2^s+(1/2*1/2^2s)+(1/3*1/2^3s)+・・+1/3^s+(1/2*1/3^2s)+(1/3*1/3^3s)+・・1/5^s+(1/2*1/5^2s)+(1/3*1/5^3s)+・・・ Then, you pick up (1/2*1/3^2s) from it. I don't know the reason, but I follow Prime obsession . I can't understand this process, but this leads to J function You may see a form which is almost a rectangular (width="3^2→∞", height="1/2"). J(x) include ∞, and 1/2*s*∫[3^2→∞]x^(-s-1)dx is also ∞. Is this like a black hole?

Riemann's zeta function is expressed like this. This depends on Sieve of Eratosthenes. According to index principle, log1/A=－logA,so Isaac Newton found x=1/p^s (p is prime) ∴ This pattern is really complicated and miraculous. I'm in the spiral road. Then, you need infinitesimal calculus. I referred to Prime obsession .

I counted prime numbers up to 100000000 by using visual basic. The graph is like this.Prime numbers increase infinitely. However, when you add more integral numbers, there are less prime numbers. This is known as power law.

I programmed to count prime numbers by PC, so I can see the same pattern from " Prime Obsession ". There is the function about prime numbers. π means the function of prime numbers up to x. For example, you can put 100 to x. 10 is the square root of 100, 4.6415・・is the cube root of 100, 3.1622・・is the fourth root of 100・・,2.1544・・is the sixth root of 100,and 1.9306・・is the seventh root of 100. 2 is the first prime number,so you can ignore the seventh root of 100. Therefore J(100)=π(100)+1/2π(10)+1/3π(4.64・・)+1/4π(3.16・・)+1/5π(2.51・・)+1/6π(2.15・・)+0+0・・・ You can count prime numbers by 100, so you see π(100)=25, π(10)=4, π(4.64・・)=2, π(3.16・・）=2, π(2.51・・)=1, π(2.15・・)=1. J(100)=25+(1/2×4)+(1/3×2)+(1/4×2)+(1/5×1)+(1/6×1) This is the graph. X is a prime number. When x is 2, J(2)=1. When x is 3, J(3) jump to 2. When x is 5, J(5) jump to 3.5. This graph would go up infinitely.

I found the pattern of prime numbers. It must be ridiculous but it seems to be almost perfect as long as I know. The realm includes infinity, so it is impossible for me to prove it. However, it keeps moving with the same pattern. You may say that it must be a fractal which means that a part includes everything. If you are interested in a long spiral road, read this blog. I will extend a stupid story from prime numbers to the universe. ρ must be optional odd numbers. Ш is apparent prime numbers such as 2,3,5. In this case, you can ignore 2 because only odd numbers must be the target to expand prime numbers. If ρ is divided by Ш and you can get the integral answer, it is Шn^2. It must be Composite numbers before Шn^2 such as 15=5*3. These are not prime numbers. ∴ For example, you can find prime numbers up to 100 by following this theory. 《３、５、７、９、１１、１３，１５、１７、１９、２１、２３、２５、２７、２９、３１、３３、３５、３７、３９、４１、４３、４５、４７、４９、５１、５３、５５、５７、５９、６１、６３、６５、６７、６９、７１、７３、７５、７７、７９、８１、８３、８５、８７、８９、９１、９３、...